4.7 Introduction to Block Matrix

Block matrix of diagonalizable linear map

Let \(\dim V=n,f:V\to V\) be \(k\)-linear diagonalizable. Let \(c_1,...,c_s\) be eigenvalues \(c_i\neq c_j\), if \(i\neq j\)

We know \(V=V_{c_1}\oplus V_{c_2}\oplus...\oplus V_{c_s},\forall i\) and \(V_{c_i}\) is \(f\)-invariant, then \(v_i\neq 0_v\) and \(v_i\in V_{c_i}\), then \(V_{c_i}\supseteq \lang v_i\rang=\lang v_i\rang_f\) and \(m_{f,v_i}(x)=x-c_i\)

And \(x-c_{i}=m_{f|_{v_i}}(x)\mid m_{f}(x)\) for all \(i\), this isn't the decomposition by the cyclic dec thm because the minimal polynomials are coprime (contrapositive of theorem)

Example

\(f:\R^3\to \R^3\) and \(f\) is diagonalizable. \([f]_{E}= \begin{pmatrix} 2 & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} \\ \placeholder{} & \placeholder{} & 3 \end{pmatrix}\)​

\(\R^3=V_1\oplus V_2\) where \(V_1\) is eigenvectors of eigenvalue 2 and \(V_2\) is eigenvectors of eigenvalue 3

Then \(\R^{3}=V_{1}\oplus V_{2}=\lang e_1+e_2\rang\oplus \lang e_3\rang\)

Then \(\chi_f(x)=(x-2)^2(x-3)\) and \(m_f(x)=(x-2)(x-3)\) and \(m_{f|_{v_1}}(x)=x-2\), \(m_{f|_{v_2}}(x)=x-3\)

Then we assume that \(\R^3=\lang z_1\rang_f\oplus ...\oplus \lang z_r\rang_f\) like in the C.D.T. with \(W_{0}=\{0\}\)

Then \(m_{f,z_i}(x)\mid m_{f,z_{i-1}}(x),\forall i,2\leq i\leq r\) and \(m_{f,z_i}(x)\mid m_f(x)=(x-2)(x-3)\) where \(r\leq 3\)

Then \(m_{f,z_1}(x)=(x-2)(x-3),m_{f,z_2}(x)\mid (x-2)(x-3)\), then \(m_{f,z_2}=(x-2)(x-3)\) or \((x-2)\) or \((x-3)\)

Since basis of \(V_1\) is \(\{e_1,e_2\}\) and basis of \(V_2\) is \(\{e_{3}\}\)

Let pick \(e_1+e_3\), then \(f(e_1+e_3)=2e_1+3e_3\), then \(\lang e_{1}+e_{3}\rang_{f}=\lang e_1+e_3,2e_1+3e_3\rang\) is l.i. since they belongs to different eigenspace

Then \(f^{2}(e_{1}+e_{3})=4e_{1}+9e_{3}\in \lang e_{1}+e_{3}\rang_{f}\), then \(m_{f,e_1+e_3}(x)=\left(x-2\right)\) or \((x-3)\) or \((x-2)(x-3)\).

The first two terms are impossible since \(e_1+e_3\) are not an eigenvector

Thus \(m_{f,e_1+e_3}(x)=m_{f}(x)\)

Let \(z_1=e_1+e_3\), then \(\dim\lang z_1\rang_f=2\), so \(r=2\) and \(\dim\lang z_2\rang_f=1\), then \(z_2\) is an eigenvector

Take \(z_2=e_2\), then \(B^{\prime}=\{z_{1},f(z_{1}),z_{2}\}\). What is \(f^2(z_1)\)?

\(m_f(f)=f^2-5f+6Id=0\), then \(f^2(z_1)=5f(z_1)-6z_2\)

Then \([f]_{\mathcal{B^{\prime}}}=\)​

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General Case

\(\dim V=n\), \(f:V\to V\) is diagonalizable and \(\chi_{f}(x)=(x-c_{1})^{d_1}\cdots(x-c_{s})^{d_{s}}\) where \(d_1+...+d_s=n\)

Then \(V=V_{c_1}\oplus ...\oplus V_{c_s}\) where \(c_1,...,c_s\) are different eigenvalues

Let basis for \(V_{c_1}\) be \(B_{1}=\{w_{1}^{1},...,w_{d_1}^{1}\}\), in general, for each \(i, 1 \leq i \leq s\) let \(B_{i}=\{w_{1}^{i},w_{2}^{i},...,w_{d_{i}}^{i}\}\) be an ordered basis of \(V_{c_i}\).This \(i\) is index not power

Define vectors \(z_1, ..., z_l \in V\) by: given \(1 \leq j \leq l\), \(z_{j}=\sum_{i/1\leq j\leq d_{i}}w_{j}^{i}\)

where \(i\) vary all the possibilities and \(j\) can be fixed, or choose any one in \([1,d_i]\)(don't repeat)

Exercise

1. Prove that \(\langle z_{j}\rangle_{f}=\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}\)

   Proof

   Assume \(\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}=\langle w_{j}^{i_1},w_{j}^{i_2},. ..,w_{j}^{i_{k}}\rangle\) where \(i_1\neq i_2\neq ...\neq i_k\), then \(\dim\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}=k\)​

   Also since \(z_{j}=w_{j}^{i_1}+w_{j}^{i_2}+...+w_{j}^{i_k}\), \(f(z_{j})=c_{i_1}w_{j}^{i_1}+c_{i_2}w_{j}^{i_2}+...+c_{i_k}w_{j}^{i_{k}}\),..., \(f^{m}(z_{j})=c_{i_1}^{m}w_{j}^{i_1}+c_{i_2}^{m}w_{j}^{i_2}+...+c_{i_{k}}^{m}w_{j} ^{i_{k}}\)

   Then any vector in \(\langle z_{j}\rangle_{f}\) is a linear combination of \(z_j,...,f^m(z_j)\), then clearly it also in \(\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}\)​

   Thus \(\langle z_{j}\rangle_{f}\subseteq\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}\)

   Then we only need to check the dimension

   We know \(\dim\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}=k\), how about \(\dim\langle z_{j}\rangle_{f}\)?

   By theorem we know \(\dim\langle z_{j}\rangle_{f}=\deg m_{f,z_{j}}(x)\)

   And we know \(w_{j}^{i_1},w_{j}^{i_2},...,w_{j}^{i_{k}}\) are in different eigenspace and \(f\) is diagonalizable

   Then \(m_{f,z_{j}}(x)=\left(x-c_{i_1}\right)\left(x-c_{i_2}\right)...(x-c_{i_{k}})\) (we use (2) here)

   Clearly, \(\deg m_{f,z_{j}}(x)=k\)

   Thus \(\dim\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}=\dim\langle z_{j}\rangle_{f}\)​

   Then \(\langle z_{j}\rangle_{f}=\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}\)

2. Prove that \(m_{f,z_j}(x) = \prod_{i / 1 \leq j \leq d_i}(x - c_{i})\)

   We know \(m_{f,z_{j}}(x)\mid m_{f}(x)\), since \(f\) is diagonalizable, than \(m_f(x)=(x-c_1)(x-c_2)...(x-c_s)\)

   Thus \(m_{f,z_{j}}(x)\) is the divisor of such linear factor

   Since \(\langle w_{j}^{i}\rangle_{i/1\leq j\leq d_{i}}=\langle w_{j}^{i_1},w_{j}^{i_2},. ..,w_{j}^{i_{k}}\rangle\), then suppose there exists \(c_{j}\neq c_{i_1},...,c_{i_k}\) such that \((x-c_{j})\mid m_{f,z_{j}}(x)\)

   But we know \((f-c_{j}Id)\left(z_{j}\right)=(f-c_{j}Id)\left(w_{j}^{i_1}\right)+\cdots+(f-c_{j}Id)\left (w_{j}^{i_{k}}\right)\neq0\) since \(c_j\) is not corresponding eigenvalue of \(w_j^{i_1},...,w_j^{j_k}\)

   Thus this contradicts to the minimality.

   Thus \(m_{f,z_j}(x) = \prod_{i / 1 \leq j \leq d_i}(x - c_{i})\)

Then \(m_{f, z_j}(x) | m_{f, z_{j-1}}(x)\), \(\forall j\)

Remark

If \(B=\{v_1,...,v_n\}\) is a basis of \(V\), then \(m_{f}(x)=\text{LCM}\{m_{f,v_{i}}(x)\}_{1\leq i\leq n}\)

Proof

• \(\text{LCM}\{m_{f,v_i}(x)\}=p(x)\) is monic
• \(m_{f,v_i}(x)\mid p(x)\), \(\forall i,1\leq i\leq n\)​
• If \(m_{f,v_i}(x)\mid h(x)\), \(\forall h(x)\in k[x]\), then \(p(x)\mid h(x)\)

Take \(h(x)=m_f(x)\), we get \(p(x)\mid m_f(x)\).

Given \(v\in V\), \(\exists \lambda_1,...,\lambda_n\in k:v=\lambda_1v_1+...+\lambda_nv_n\), then \(p(f)(v)=\lambda_{1}p(f(v_{1}))+\cdots+\lambda_{n}p(f(v_{n}))\)

And \(\forall i,p(f)(v_i)=0_V\) because \(m_{f,v_i}(x)\mid p(x)\). So \(p(f)(v)=0\). Thus \(m_f(x)\mid p(x)\)

Then \(p(x)=m_{f}(x)\)​

Comment: Sometimes it's easier to use find vector minimal polynomial then find minimal polynomial

Example

\(\dim V=n,0\neq N:V\to V\) is nilpotent linear map

So \(N\) is not diagonalizable, then \(\chi_N(x)=x^n\), \(m_N(x)=x^i\) where \(i\) is the index of nilpotency and \(1\leq i\leq n\)

Using CDT for \(W_0=\{0_v\}\), we know \(\exists z_{1},...,z_{r}\in V:V=\langle z_{1}\rangle_{N}\oplus...\oplus\langle z_{r} \rangle_{N}\) and \(m_{N,z_i}(x)\mid m_{N,z_{i-1}}(x)\) \(\forall i,2\leq i\leq r\)

Then \(m_{N,z_{j}}(x)=x^{l_{j}}\) for some \(l_j\) and \(1\leq l_j\leq i\) since \(m_{N,z_{j}}(x)\mid m_{N}(x)\)

Then we know \(i=l_{1}\geq l_{2}\geq...\geq l_{r}\) since \(m_{f}(x)=\text{LCM}\{m_{f,v_{i}}(x)\}_{1\leq i\leq n}\)

Also \(\dim l_{1}=i,...,\dim l_{r}=l_{r}\). Then the basis of \(\lang z_{1}\rang_{N}\) is \(B_{1}=\{z_{1},N\left(z_{1}\right),\ldots,N^{l_1-1}(z_1)\}\)

...... the basis of \(\lang z_{r}\rang_{N}\) is \(B_{r}=\{z_{r},N\left(z_{r}\right),\ldots,N^{l_r-1}(z_{r})\}\)

Then \(B=B_1\cup ...\cup B_r\) is basis for \(V\)

Thus \(l_{1}=\dim\langle z_{1}\rangle_{N}\geq\dim\langle z_{2}\rangle_{N}\geq\ldots\geq \dim\langle z_{r}\rangle_{N}\), then the sizes of \(A_i\)'s are non increasing

​

The size of \(A_1\) is \(i\times i\) where \(i=\deg m_N(x)\)

\(\forall j\), \(A_{j}=[N|_{\lang z_{j}\rang_N}]_{B_j}=\)​

Then we have \(r\) columns of \(0's\) in \([N]_B\), then \(\dim \ker N=r\)

Then a basis of \(\ker N=\{N^{l_1-1}(z_1),N^{l_2-1}(z_2),...,N^{l_r-1}(z_r)\}\)

Example

\(f:\mathbb{C}^4\to \mathbb{C}^4\) and \(\mathbb{C}\)-linear, \([f]_{E}= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)

\(\chi_{f}(x)=x^{4}\) and \(f\) is nilpotent. Then \([f]^2_E\neq 0\) and \([f]_E^3\neq 0\), so \(m_f(x)=x^4\)

Use CDT: \(\mathbb{C}^4=\lang z_1\rang_f\) since \(\dim\lang z_1\rang _f=\deg m_f(x)=4\)

Then choose \(z_1\), basis \(B=\{z_{1},f(z_{1}),f^{2}(z_{1}),f^{3}(z_{1})\}\) so \(z_{1}\in\ker f^{4}\setminus\ker f^{3}\), \(f(z_1)\in \ker f^3\setminus \ker f^2\),..., \(f^{3}(z_{1})\in \ker f\setminus \{0_{V}\}\)

Take \(z_1=e_4\) since \(e_{4}\mapsto3e_{3}\mapsto6e_{2}\mapsto6e_{1}\mapsto0\) by \(f\)

\([f]_{B}= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}\) and \(\dim \ker f=1,\ker f=\lang e_1\rang\)

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Now let \(N:V\to V,\dim V=n\) is a nilpotent linear map, let \(\lambda \in k\), \(f=N+\lambda Id\)

\(\chi_N(x)=x^n,m_N(x)=x^i\) where \(i\) is the index of nilpotency

\(0=N^i=(f-Id)^i\) and \(0\neq N^{i-1}\)

\(\chi_f(x)=(x-\lambda)^n\) and \(m_f(x)=(x-\lambda)^i\)

We know what to do with \(N\)(using the CDT)

\(\exists z_{1},...,z_{r}:V=\langle z_{1}\rangle_{N}\oplus...\oplus\langle z_{r}\rangle _{N}\) and since \(r\) is the dimension of kernel, then \(r=\dim \ker N=\dim \ker(f-\lambda Id)\)

Building a basis \(B=B_1\cup ...\cup B_r\) as before,​

Then \([f]_B=[N+\lambda Id]_B=[N]_B+\lambda [Id]_B\)

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